7 Sample questionA stone block with a mass of 800Kg is lifted from rest with a uniform acceleration by a crane such that it reached a velocity of 14m/s. after 10 seconds. Calculate the tension in the lifting cable

11 Force due to accelerationSample questionIf the stone is lowered at the same rate of acceleration the tension is less than the weight of the boxForce due to accelerationF = maTensionaccelerationWeight mgTension in cable == 6728N (6.7kN)

12 LiftYou get the same effect when you travel in a lift. You feel heavier when the lift begins and accelerates upwards and lighter when the lift accelerates downwards

13 A man standing on weighing scales in a lift has a mass of 60 KgA man standing on weighing scales in a lift has a mass of 60 Kg. The lift accelerates uniformly at a rate of 2m/s2 Calculate the reading on the scales during the period of acceleration .

29 D’Alembert’s PrincipleWhen an object is accelerating the applied force making it accelerate has to overcome the inertia. This is the force which resists the acceleration (or deceleration) and is equal and opposite to the applied force. This means that the total force acting on the body is zeroAcceleration (a)Applied Force (F)Inertia Force (Fi)F + Fi = 0 so F = -FiF = maso ma = - Fior Fi = - mamass

34 Kinetic energyGPE = mgh = joulesNeglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottom5Kg10mKE (at the bottom) =490.5J = mv22(v = velocity)

35 The velocity at the bottom v =√(2Ke÷m)Kinetic energyNeglecting friction, all the GPE at the top of the slope converts to kinetic energy at the bottomGPE = mgh = joules5KgThe velocity at the bottom v =√(2Ke÷m)= √ (2 x ÷ 5)√196.2= 14m/s10m

37 ExampleA car of mass 800Kg is stood on a uniform 1 in 10 slope when it’s handbrake is suddenly released and it runs 30 metres to the bottom of the slope against a uniform frictional force of 50N. What is the car’s velocity at the bottom of the slope?

44 Example 2 (Resolving forces method)Work out the forces involvedExample 2 (Resolving forces method)Wt = weight of car (7848N)Fn is the normal reaction force of the slope on the carFs is the force on the car down the slope5.7oFnFswt

50 Momentum = mass x velocity The total momentum remains the same before and after a collisionMomentum is a vector quantity

51 Linear Momentum and CollisionsA railway coach of mass 25t is moving along a level track 36km/hr when it collides with and couples up to another coach of mass 20t moving in the same direction at 6km/hr. Both of the coaches continue in the same direction after coupling. What is the combined velocity of the two coaches?

52 Linear Momentum and CollisionsLet the mass of the first coach be M1 and the mass of the second coach be M2 and the velocity of the first coach be V1 and the velocity of the second be V2

53 Linear Momentum and CollisionsBefore coupling the momentum of the first coach is 25 x 36 = 900tkm/hr and the momentum of the second is 20 x 6 =120 tkm/hrWhich is a total of 1020tkm/hr

54 Linear Momentum and CollisionsAfter the coupling the momentum of both is the same as before the coupling which is tkm/hrAnd the combined mass is 45t

55 Linear Momentum and CollisionsVelocity after coupling is momentum divided by mass1020÷4522.6 km/hr

56 ExampleA hammer of mass 200Kg falls 5m on to a pile of mass 300Kg and drives it 100mm into the grounda) Calculate the loss of energy on the impact.b) Calculate the work done by the resistance of the ground.c) calculate the average resistance to penetration.

61 Work done by ground = Force x distanceExampleWork done by ground = Force x distance39.2kN x 0.1 m3.92kj(this agrees with the fact that fact that the kinetic energy of the hammer and pile after impact was 3.92kj and zero when the pile stopped moving in the ground